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• 1 指数和对数映射

1 指数和对数映射

对李代数拆分，$\varphi =\theta a$，模长$\theta$，方向为单位向量$a,||a||=1$，有以下性质：
a ∧ a ∧ = a a T − I a ∧ a ∧ a ∧ = a ∧ ( a a T − I ) = − a ∧ a^{\land}a^{\land} = aa^T-I \\ a^{\land}a^{\land}a^{\land}=a^{\land}(aa^T-I)=-a^{\land} a∧a∧=aaT−Ia∧a∧a∧=a∧(aaT−I)=−a∧
对于$so(3)$的任意向量$\varphi$，可以定义指数映射，以及对应的泰勒展开：
c o s x = 1 − x 2 2 ! + x 4 4 ! − x 6 6 ! + ⋯ s i n x = x − x 3 3 ! + x 5 5 ! − x 7 7 ! + ⋯ R = e ϕ ∧ = ∑ n = 0 1 n ! ( ϕ ∧ ) n = ∑ n = 0 1 n ! ( θ a ∧ ) n = I + θ a ∧ + 1 2 ! θ 2 a ∧ a ∧ + 1 3 ! θ 3 a ∧ a ∧ a ∧ + 1 4 ! θ 4 a ∧ a ∧ a ∧ a ∧ + ⋯ = a a T − a ∧ a ∧ + θ a ∧ + 1 2 ! θ 2 a ∧ a ∧ − 1 3 ! θ 3 a ∧ − 1 4 ! θ 4 a ∧ a ∧ + ⋯ = a a T + ( θ − 1 3 ! θ 3 + 1 5 ! θ 5 − ⋯ ) a ∧ − ( 1 − 1 2 ! θ 2 + 1 4 ! θ 4 − ⋯ ) a ∧ a ∧ = a ∧ a ∧ + I + s i n θ a ∧ − c o s θ a ∧ a ∧ = ( 1 − c o s θ ) a ∧ a ∧ + I + s i n θ a ∧ = c o s θ I + ( 1 − c o s θ ) a a T + s i n θ a ∧ cosx = 1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+ \cdots \\ sinx = x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+ \cdots \\ R = e^{\phi^{\land}}= \sum_{n=0}{\frac{1}{n!}(\phi^{\land})^n} = \sum_{n=0}{\frac{1}{n!}(\theta a^{\land})^n} \\ = I + \theta a^{\land} + \frac{1}{2!}\theta^2a^{\land}a^{\land} + \frac{1}{3!}\theta^3a^{\land}a^{\land}a^{\land} + \frac{1}{4!}\theta^4a^{\land}a^{\land}a^{\land}a^{\land} + \cdots \\ = aa^T - a^{\land}a^{\land} + \theta a^{\land} + \frac{1}{2!}\theta^2a^{\land}a^{\land} -\frac{1}{3!}\theta^3a^{\land} - \frac{1}{4!}\theta^4a^{\land}a^{\land} + \cdots \\ = aa^T + (\theta - \frac{1}{3!}\theta^3 + \frac{1}{5!}\theta^5 - \cdots)a^{\land} - (1-\frac{1}{2!}\theta^2+\frac{1}{4!}\theta^4-\cdots)a^{\land}a^{\land} \\ = a^{\land}a^{\land} + I + sin\theta a^{\land} - cos\theta a^{\land}a^{\land} \\ = (1-cos\theta)a^{\land}a^{\land} +I +sin\theta a^{\land} \\ = cos\theta I + (1-cos\theta)aa^T+sin\theta a^{\land} cosx=1−2!x2​+4!x4​−6!x6​+⋯sinx=x−3!x3​+5!x5​−7!x7​+⋯R=eϕ∧=n=0∑​n!1​(ϕ∧)n=n=0∑​n!1​(θa∧)n=I+θa∧+2!1​θ2a∧a∧+3!1​θ3a∧a∧a∧+4!1​θ4a∧a∧a∧a∧+⋯=aaT−a∧a∧+θa∧+2!1​θ2a∧a∧−3!1​θ3a∧−4!1​θ4a∧a∧+⋯=aaT+(θ−3!1​θ3+5!1​θ5−⋯)a∧−(1−2!1​θ2+4!1​θ4−⋯)a∧a∧=a∧a∧+I+sinθa∧−cosθa∧a∧=(1−cosθ)a∧a∧+I+sinθa∧=cosθI+(1−cosθ)aaT+sinθa∧
可见$so(3)$就是旋转向量组成的空间，而指数映射就是罗德里格斯公式。而对数映射可以将李群转为李代数 ϕ = θ a = l n ( R ) ∨ \ \phi=\theta a=ln(R)^{\lor}  ϕ=θa=ln(R)∨，但计算较为复杂，可使用以下公式进行计算角度，根据转轴不变求解方向向量：

R = c o s θ I + ( 1 − c o s θ ) a a T + s i n θ a ∧ t r ( R ) = c o s θ t r ( I ) + ( 1 − c o s θ ) t r ( a a T ) + s i n θ t r ( a ∧ ) = 3 c o s θ + ( 1 − c o s θ ) = 1 + 2 c o s θ θ = a r c c o s t r ( R ) − 1 2 R a = a R = cos\theta I + (1-cos\theta)aa^T+sin\theta a^{\land} \\ tr(R) = cos\theta tr(I) + (1-cos\theta)tr(aa^T)+sin\theta tr(a^{\land}) \\ = 3cos\theta+(1-cos\theta) = 1+2cos\theta\\ \theta = arccos \frac{tr(R)-1}{2} \\ Ra = a R=cosθI+(1−cosθ)aaT+sinθa∧tr(R)=cosθtr(I)+(1−cosθ)tr(aaT)+sinθtr(a∧)=3cosθ+(1−cosθ)=1+2cosθθ=arccos2tr(R)−1​Ra=a
对于$se(3)$进行指数映射，推导如下：
T = e ϵ ∧ = ∑ n = 0 1 n ! ( ϵ ∧ ) n = I + ∑ n = 1 1 n ! ( [ ϕ ∧ ρ 0 T 0 ] ) n T = e^{\epsilon^{\land}} = \sum_{n=0}{\frac{1}{n!}(\epsilon^{\land})^n} \\ = I + \sum_{n=1}{\frac{1}{n!}(\begin{bmatrix} \phi^{\land} & \rho \\ 0^T & 0 \end{bmatrix})^n} \\ T=eϵ∧=n=0∑​n!1​(ϵ∧)n=I+n=1∑​n!1​([ϕ∧0T​ρ0​])n
对矩阵乘法结果进行归纳后得到：
[ θ a ∧ ρ 0 T 0 ] [ θ a ∧ ρ 0 T 0 ] = [ ( θ a ∧ ) 2 θ a ∧ ρ 0 T 0 ] \begin{bmatrix} \theta a^{\land} & \rho \\ 0^T & 0 \end{bmatrix} \begin{bmatrix} \theta a^{\land} & \rho \\ 0^T & 0 \end{bmatrix} = \begin{bmatrix} (\theta a^{\land})^2 & \theta a^{\land}\rho \\ 0^T & 0 \end{bmatrix} \\ [θa∧0T​ρ0​][θa∧0T​ρ0​]=[(θa∧)20T​θa∧ρ0​]
最终指数映射结果如下：
T = I + ∑ n = 1 1 n ! [ ( θ a ∧ ) n ( θ a ∧ ) n − 1 ρ 0 T 0 ] = [ ∑ n = 0 1 n ! ( ϕ ∧ ) n ∑ n = 1 1 n ! ( ϕ ∧ ) n − 1 ρ 0 T 1 ] = [ ∑ 1 n ! ( ϕ ∧ ) n ∑ 1 ( n + 1 ) ! ( ϕ ∧ ) n ρ 0 T 1 ] = [ R J ρ 0 T 1 ] T = I + \sum_{n=1}{\frac{1}{n!}\begin{bmatrix} (\theta a^{\land})^n & (\theta a^{\land})^{n-1}\rho \\ 0^T & 0 \end{bmatrix}} \\ = \begin{bmatrix} \sum_{n=0}{\frac{1}{n!}(\phi^{\land})^n} & \sum_{n=1}{\frac{1}{n!}(\phi^{\land})^{n-1}}\rho \\ 0^T & 1 \end{bmatrix} \\ = \begin{bmatrix} \sum{\frac{1}{n!}(\phi^{\land})^n} & \sum{\frac{1}{(n+1)!}(\phi^{\land})^n}\rho \\ 0^T & 1 \end{bmatrix} = \begin{bmatrix} R & J\rho \\ 0^T & 1 \end{bmatrix} T=I+n=1∑​n!1​[(θa∧)n0T​(θa∧)n−1ρ0​]=[∑n=0​n!1​(ϕ∧)n0T​∑n=1​n!1​(ϕ∧)n−1ρ1​]=[∑n!1​(ϕ∧)n0T​∑(n+1)!1​(ϕ∧)nρ1​]=[R0T​Jρ1​]
和$so(3)$一样将R转为$\varphi =\theta a$的推导与so(3)映射一致，
J = ∑ n = 0 1 ( n + 1 ) ! ( ϕ ∧ ) n = ∑ n = 0 1 ( n + 1 ) ! ( θ a ∧ ) n = I + 1 2 ! θ a ∧ + 1 3 ! θ 2 a ∧ a ∧ + 1 4 ! θ 3 a ∧ a ∧ a ∧ + 1 5 ! θ 4 a ∧ a ∧ a ∧ a ∧ + ⋯ = 1 θ ( 1 2 ! θ 2 − 1 4 ! θ 4 + ⋯ ) a ∧ + 1 θ ( 1 3 ! θ 3 − 1 5 ! θ 5 + ⋯ ) a ∧ a ∧ + I = 1 θ ( 1 − c o s θ ) a ∧ + 1 θ ( θ − s i n θ ) a ∧ a ∧ + I = 1 − c o s θ θ a ∧ + 1 θ ( θ − s i n θ ) ( a a T − I ) + I = 1 − c o s θ θ a ∧ + ( 1 − s i n θ θ ) ( a a T − I ) + I = 1 − c o s θ θ a ∧ + ( 1 − s i n θ θ ) a a T + s i n θ θ I t = J ρ J = \sum_{n=0}{\frac{1}{(n+1)!}(\phi^{\land})^n} = \sum_{n=0}{\frac{1}{(n+1)!}(\theta a^{\land})^n} \\ = I + \frac{1}{2!}\theta a^{\land} + \frac{1}{3!}\theta^2a^{\land}a^{\land} + \frac{1}{4!}\theta^3a^{\land}a^{\land}a^{\land} + \frac{1}{5!}\theta^4a^{\land}a^{\land}a^{\land}a^{\land} + \cdots \\ = \frac{1}{\theta}(\frac{1}{2!}\theta^2-\frac{1}{4!}\theta^4+\cdots)a^{\land}+\frac{1}{\theta}(\frac{1}{3!}\theta^3-\frac{1}{5!}\theta^5+\cdots)a^{\land}a^{\land}+I \\ = \frac{1}{\theta}(1-cos\theta)a^{\land}+\frac{1}{\theta}(\theta-sin\theta)a^{\land}a^{\land}+I \\ = \frac{1-cos\theta}{\theta}a^{\land}+\frac{1}{\theta}(\theta-sin\theta)(aa^T - I)+I \\ = \frac{1-cos\theta}{\theta}a^{\land} + (1-\frac{sin\theta}{\theta})(aa^T - I) + I \\ = \frac{1-cos\theta}{\theta}a^{\land} + (1-\frac{sin\theta}{\theta})aa^T + \frac{sin\theta}{\theta}I \\ t = J\rho J=n=0∑​(n+1)!1​(ϕ∧)n=n=0∑​(n+1)!1​(θa∧)n=I+2!1​θa∧+3!1​θ2a∧a∧+4!1​θ3a∧a∧a∧+5!1​θ4a∧a∧a∧a∧+⋯=θ1​(2!1​θ2−4!1​θ4+⋯)a∧+θ1​(3!1​θ3−5!1​θ5+⋯)a∧a∧+I=θ1​(1−cosθ)a∧+θ1​(θ−sinθ)a∧a∧+I=θ1−cosθ​a∧+θ1​(θ−sinθ)(aaT−I)+I=θ1−cosθ​a∧+(1−θsinθ​)(aaT−I)+I=θ1−cosθ​a∧+(1−θsinθ​)aaT+θsinθ​It=Jρ
同样对数映射较为复杂，和$SO(3)$一样的方式根据R求对应的旋转向量$\varphi =\theta a$，同时根据$t=J\rho$求线性方程得到$\rho$。

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